Question: The equation of a circle $C$ is $x^2+y^2+18x-16y+129 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+18x) + (y^2-16y) = -129$ $(x^2+18x+81) + (y^2-16y+64) = -129 + 81 + 64$ $(x+9)^{2} + (y-8)^{2} = 16 = 4^2$ Thus, $(h, k) = (-9, 8)$ and $r = 4$.